EQUATIONS INVOLVING FRACTIONS (RATIONAL EQUATIONS)


Note:




If you would like an in-depth review of fractions, click on Fractions.



Solve for x in the following equation.


Example 1: $\displaystyle \frac{x^{2}-8}{x^{2}-4}+\displaystyle \frac{2}{x+2}=\displaystyle \frac{5}{x-2}$



Rewrite the equation such that all the denominators are factored:

\begin{eqnarray*}&& \\
\displaystyle \frac{x^{2}-8}{\left( x-2\right) \left( x+...
...isplaystyle \frac{2}{x+2} &=&
\displaystyle \frac{5}{x-2} \\
&&
\end{eqnarray*}


Recall that you cannot divide by zero. Therefore, the first fraction is valid if , $\quad x\neq \pm 2,$    the second fraction is valid if      $
x\neq -2,\quad $and the third fraction is valid is    $x\neq 2$.    If either $\quad 2$     or    $-2\quad $ turn out to be the solutions, you must discard them as extraneous solutions.




Multiply both sides of the equation by an expression that represents the lowest common denominator. The expression $\left( x-2\right) \left(
x+2\right) $ is the smallest expression because it is the smallest expression that is divisible by all three denominators.




\begin{eqnarray*}\displaystyle \frac{x^{2}-8}{\left( x-2\right) \left( x+2\right) }+\displaystyle \frac{2}{x+2} &=&
\displaystyle \frac{5}{x-2}
\end{eqnarray*}



\begin{eqnarray*}\left( x-2\right) \left( x+2\right) \cdot \left( \displaystyle ...
...left( x+2\right) \cdot \left( \displaystyle \frac{5}{x-2}\right)
\end{eqnarray*}



\begin{eqnarray*}\left( x-2\right) \left( x+2\right) \cdot \left( \displaystyle ...
...left( x+2\right) \cdot \left( \displaystyle \frac{5}{x-2}\right)
\end{eqnarray*}





This equation can be written as

\begin{eqnarray*}\frac{\left( x-2\right) \left( x+2\right) }{1}\cdot \displaysty...
...\right) \left( x+2\right) }{1}\cdot
\displaystyle \frac{5}{x-2}
\end{eqnarray*}




Multiply the fractions where indicated.

\begin{eqnarray*}\frac{\left( x-2\right) \left( x+2\right) \left( x^{2}-8\right)...
...2\right) \left(
x+2\right) \left( 5\right) }{\left( x-2\right) }
\end{eqnarray*}




Rearrange the factors in the numerators and rewrite the equations as

\begin{eqnarray*}\frac{\left( x-2\right) \left( x+2\right) \left( x^{2}-8\right)...
...2\right) \left(
x+2\right) \left( 5\right) }{\left( x-2\right) }
\end{eqnarray*}




Rewrite the equation once again as


\begin{eqnarray*}\frac{\left( x-2\right) \left( x+2\right) }{\left( x-2\right) \...
...( x-2\right) }\cdot \frac{\left( x+2\right)
\left( 5\right) }{1}
\end{eqnarray*}



\begin{eqnarray*}1\cdot \frac{\left( x^{2}-8\right) }{1}+1\cdot \frac{\left( x-2...
... x-2\right) \left( 2\right) &=&\left(
x+2\right) \left( 5\right)
\end{eqnarray*}




Simplify the last equation and solve for x.

\begin{eqnarray*}x^{2}-8+2x-4 &=&5x+10 \\
&& \\
&& \\
x^{2}+2x-12 &=&5x+10 \\
&& \\
&& \\
x^{2}-3x-22 &=&0
\end{eqnarray*}



\begin{eqnarray*}x &=&\frac{-\left( -3\right) \pm \sqrt{\left( -3\right) ^{2}-4\left(
1\right) \left( -22\right) }}{2\left( 1\right) }
\end{eqnarray*}



\begin{eqnarray*}x &=&\frac{3\pm \sqrt{97}}{2} \\
&& \\
&& \\
x &=&\frac{3}{2...
...&& \\
x &=&\frac{3}{2}-\frac{\sqrt{97}}{2}\approx -3.4244289009
\end{eqnarray*}


The original equation has two real solutions: $x=\displaystyle \frac{3}{2}+\displaystyle \frac{\sqrt{97
}}{2}\smallskip\approx...
... \frac{3}{2}-\displaystyle \frac{
\sqrt{97}}{2}\smallskip\approx -3.4244289009.$ The exact answers are $x=
\displaystyle \frac{3}{2}\pm \displaystyle \frac{\sqrt{97}}{2}$ and the approximate answers are $
x\approx 6.4244289009$ and -3.4244289009. In most cases it is easier to check your solutions with the approximate answers.




Check the two answers in the original equation.


Check the solution $x=\displaystyle \frac{3}{2}+\displaystyle \frac{\sqrt{97}}{2}\smallskip\approx
6.4244289009$ by substituting 6.4244289009 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.        




Since the left side of the original equation is not equal to the right side of the original equation after we substitute the value 6.4244289009 for x, then x=6.4244289009 is a solution.





Check the solution $x=\displaystyle \frac{3}{2}-\displaystyle \frac{\sqrt{97}}{2}\smallskip\approx
-3.4244289009$ by substituting -3.4244289009 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.        



Since the left side of the original equation is not equal to the right side of the original equation after we substitute the value -3.4244289009 for x, then x=-3.4244289009 is a solution.





You can also check your answer by graphing $\quad f(x)=\displaystyle \frac{x^{2}-8}{
x^{2}-4}+\displaystyle \frac{2}{x+2}-\displaystyle \frac{5}{x-2}\smallskip .\quad $(formed by subtracting the right side of the original equation from the left side). Look to see where the graph crosses the x-axis; that will be the real solution. Note that the graph crosses the x-axis in two places: -3.4244289009 and 6.4244289009.



This means that there are two real solutions and the solutions are x=-3.4244289009 and 6.4244289009.







If you would like to work another example, click on Example


If you would like to test yourself by working some problems similar to this example, click on Problem


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Author: Nancy Marcus

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